Answer
$\color{blue}{\left\{1-2i, 1+2i\right\}}$
Work Step by Step
Subtract $2x-5$ to both sides of the equation to obtain:
$x^2-(2x-5)=2x-5-(2x-5)
\\x^2-2x-(-5)=0
\\x^2-2x+5=0$
RECALL:
The quadratic equation $ax^2+bx+c=0$ can be solved using the quadratic formula:
$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
The given equation has:
$a=1, b=-2 c=5$
Substitute these values into the quadratic formula to obtain:
$x=\dfrac{-(-2) \pm \sqrt{(-2)^2-4(1)(5)}}{2(1)}
\\x=\dfrac{2\pm\sqrt{4-20}}{2}
\\x=\dfrac{2\pm\sqrt{-16}}{2}
\\x=\dfrac{2\pm\sqrt{16(-1)}}{2}
\\x=\dfrac{2\pm 4i}{2}
\\x=1 \pm 2i$
Thus, the solution set is $\color{blue}{\left\{1-2i, 1+2i\right\}}$.
